Question

One end of massless spring of spring constant $100 \,N/m$ and natural length $0.49\, m$ is fixed and other end is connected to a body of mass $0.5 \,kg$ lying on a frictionless horizontal table. The spring remains horizontal. If the body is made to rotate at an angular velocity of $2$ rad/s, then the elongation of the spring will be

• A. 2 cm
• B. 1 cm
• C. 0.5 cm
• D. 0.25 cm

Answer 1

Answer: (B)

1 cm

Answer 2

Given, $M=0.5\,kg$,
$\omega=2\, rad/ s$,
$l=0.49\, m$, and $k=100 \,N/ m$
Figure represents situation, as given in question

Centripetal force on the blocks = Spring force
$\Rightarrow Mr\omega^{2}=k \Delta x$
$\Rightarrow 0.5\left(l+\Delta x\right)\left(2\right)^{2}=100\cdot\Delta x$
$\Rightarrow \left(0.49+\Delta x\right)4=100\cdot\Delta x\times2$
$\Rightarrow 0.49+\Delta x=\frac{200}{4} \Delta x$
$\Rightarrow 0.49+\Delta x=50\Delta x$
$\therefore 49 \Delta x=0.49$
$\Delta x=0.01\, m = 1\,cm$